trig. angle of elevation The angle of elevation of the top of a pole as seen from a point $17$ feet away from the base is double the angle of elevation as seen from a point $48$ feet further from the pole. Find the height of the pole above the level of the observer’s eyes.
After drawing a diagram I could up with the equation $17\tan{2x} = 65\tan{x}$
If I could find $x$ I could find the height. But solving the equation with wolfram alpha gives me nasty answers and makes me think I'm wrong.
 A: I used $h$ in my working to signify the height. Letting $t = \tan \theta$, you get:
$$t = \frac{h}{65}$$
and $\displaystyle \tan 2\theta = \frac{2t}{1-t^2} = \frac{h}{17}$
and you end up with the quadratic:
$$\frac{(2)(\frac{h}{65})}{1-(\frac{h}{65})^2}= \frac{h}{17}$$
which is quite easy to solve (note the obvious cancellation of $h$ from both sides at the outset).
This trig solution is actually really easy. What follows is a more involved non-trigonometric solution just to show how else this can be done.
The non-trig solution:
Bisect the $2\theta$ angle to construct a triangle similar to the larger one. Let the height of the smaller triangle from the base of the pole be $y$.
By similarity, you'll be able to deduce:
$$\frac{h}{y}= \frac{65}{17}$$
Now by angle bisector theorem,
$$\frac{h-y}{y} = \frac{\sqrt{h^2 + 17^2}}{17}$$
Using the previous result,
$$(\frac{h}{17})^2 + 1 = (\frac{65}{17} - 1)^2$$
$$(\frac{h}{17})^2 = (\frac{65}{17} - 2)(\frac{65}{17})$$
$$(\frac{h}{17})^2 = \frac{(65)(31)}{17^2}$$
giving $\displaystyle h = \sqrt{2015}$
Now, aren't you glad you're allowed to use trig? :)
